\(\int\limits^{ln\sqrt{3}}_0\frac{dx}{e^x+e^{-x}}\)
Những câu hỏi liên quan
\(\int\limits^{\frac{\pi}{3}}_0\frac{sinx}{cosx\sqrt{3+sin^2x}}dx\)
\(\int\limits^{ln8}_0\frac{e^x}{1+\sqrt{3e^x+1}}dx\)
Tính các tích phân sau :
a) \(\int\limits^{\dfrac{\pi}{4}}_0\cos2x.\cos^2xdx\)
b) \(\int\limits^1_{\dfrac{1}{2}}\dfrac{e^x}{e^{2x}-1}dx\)
c) \(\int\limits^1_0\dfrac{x+2}{x^2+2x+1}\ln\left(x+1\right)dx\)
d) \(\int\limits^{\dfrac{\pi}{4}}_0\dfrac{x\sin x+\left(x+1\right)\cos x}{x\sin x+\cos x}dx\)
a)
Ta có \(A=\int ^{\frac{\pi}{4}}_{0}\cos 2x\cos^2xdx=\frac{1}{4}\int ^{\frac{\pi}{4}}_{0}\cos 2x(\cos 2x+1)d(2x)\)
\(\Leftrightarrow A=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos x(\cos x+1)dx=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos xdx+\frac{1}{8}\int ^{\frac{\pi}{2}}_{0}(\cos 2x+1)dx\)
\(\Leftrightarrow A=\frac{1}{4}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin x+\frac{1}{16}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin 2x+\frac{1}{8}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|x=\frac{1}{4}+\frac{\pi}{16}\)
b)
\(B=\int ^{1}_{\frac{1}{2}}\frac{e^x}{e^{2x}-1}dx=\frac{1}{2}\int ^{1}_{\frac{1}{2}}\left ( \frac{1}{e^x-1}-\frac{1}{e^x+1} \right )d(e^x)\)
\(\Leftrightarrow B=\frac{1}{2}\left.\begin{matrix} 1\\ \frac{1}{2}\end{matrix}\right|\left | \frac{e^x-1}{e^x+1} \right |\approx 0.317\)
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c)
Có \(C=\int ^{1}_{0}\frac{(x+2)\ln(x+1)}{(x+1)^2}d(x+1)\).
Đặt \(x+1=t\)
\(\Rightarrow C=\int ^{2}_{1}\frac{(t+1)\ln t}{t^2}dt=\int ^{2}_{1}\frac{\ln t}{t}dt+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)
\(=\int ^{2}_{1}\ln td(\ln t)+\int ^{2}_{1}\frac{\ln t}{t^2}dt=\frac{\ln ^22}{2}+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)
Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=\frac{dt}{t^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=\frac{-1}{t}\end{matrix}\right.\Rightarrow \int ^{2}_{1}\frac{\ln t}{t^2}dt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|-\frac{\ln t+1}{t}=\frac{1}{2}-\frac{\ln 2 }{2}\)
\(\Rightarrow C=\frac{1}{2}-\frac{\ln 2}{2}+\frac{\ln ^22}{2}\)
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d)
\(D=\int ^{\frac{\pi}{4}}_{0}\frac{x\sin x+(x+1)\cos x}{x\sin x+\cos x}dx=\int ^{\frac{\pi}{4}}_{0}dx+\int ^{\frac{\pi}{4}}_{0}\frac{x\cos x}{x\sin x+\cos x}dx\)
Ta có:
\(\int ^{\frac{\pi}{4}}_{0}dx=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|x=\frac{\pi}{4}\)
\(\int ^{\frac{\pi}{4}}_{0}\frac{x\cos xdx}{x\sin x+\cos x}=\int ^{\frac{\pi}{4}}_{0}\frac{d(x\sin x+\cos x)}{x\sin x+\cos x}=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\ln |x\sin x+\cos x|\)
\(=\ln|\frac{\pi\sqrt{2}}{8}+\frac{\sqrt{2}}{2}|\)
Suy ra \(D=\frac{\pi}{4}+\ln|\frac{\pi\sqrt{2}}{8}+\frac{\sqrt{2}}{2}|\)
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Tính các tích phân sau bằng phương pháp tính tích phân từng phần :
a) \(\int\limits^{e^4}_1\sqrt{x}\ln xdx\)
b) \(\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}\dfrac{xdx}{\sin^2x}\)
c) \(\int\limits^{\pi}_0\left(\pi-x\right)\sin xdx\)
d) \(\int\limits^0_{-1}\left(2x+3\right)e^{-x}dx\)
Hãy chỉ ra kết quả nào dưới đây đúng :
a) intlimits^{dfrac{pi}{2}}_0sin xdx+intlimits^{dfrac{3pi}{2}}_{dfrac{pi}{2}}sin xdx+intlimits^{2pi}_{dfrac{3pi}{2}}sin xdx0
b) intlimits^{dfrac{pi}{2}}_0left(sqrt[3]{sin x}-sqrt[3]{cos x}right)dx0
c) intlimits^{dfrac{1}{2}}_{-dfrac{1}{2}}lndfrac{1-x}{1+x}dx0
d) intlimits^2_0left(dfrac{1}{1+x+x^2+x^3}+1right)dx0
Đọc tiếp
Hãy chỉ ra kết quả nào dưới đây đúng :
a) \(\int\limits^{\dfrac{\pi}{2}}_0\sin xdx+\int\limits^{\dfrac{3\pi}{2}}_{\dfrac{\pi}{2}}\sin xdx+\int\limits^{2\pi}_{\dfrac{3\pi}{2}}\sin xdx=0\)
b) \(\int\limits^{\dfrac{\pi}{2}}_0\left(\sqrt[3]{\sin x}-\sqrt[3]{\cos x}\right)dx=0\)
c) \(\int\limits^{\dfrac{1}{2}}_{-\dfrac{1}{2}}\ln\dfrac{1-x}{1+x}dx=0\)
d) \(\int\limits^2_0\left(\dfrac{1}{1+x+x^2+x^3}+1\right)dx=0\)
Tính các tích phân sau
1.Iintlimits^{frac{Pi}{4}}_0 (x+1)sin2xdx
2.Iintlimits^2_1frac{x^2+3x+1}{x^2+x}dx
3.Iintlimits^2_1frac{x^2-1}{x^2}lnxdx
4. Iintlimits^1_0xsqrt{2-x^2}dx
5.Iintlimits^1_0frac{left(x+1right)^2}{x^2+1}dx
6. Iintlimits^5_1frac{dx}{1+sqrt{2x-1}}
7. Iintlimits^3_1frac{1+lnleft(x+1right)}{x^2}dx
8.Iintlimits^1_0frac{x^3}{x^4+3x^2+2}dx
9. Iintlimits^{frac{Pi}{4}}_0xleft(1+sin2xright)dx
10. Iintlimits^3_0frac{x}{sqrt{x+1}}dx
Đọc tiếp
Tính các tích phân sau
1.I=\(\int\limits^{\frac{\Pi}{4}}_0\) (x+1)sin2xdx
2.I=\(\int\limits^2_1\frac{x^2+3x+1}{x^2+x}dx\)
3.I=\(\int\limits^2_1\frac{x^2-1}{x^2}lnxdx\)
4. I=\(\int\limits^1_0x\sqrt{2-x^2}dx\)
5.I=\(\int\limits^1_0\frac{\left(x+1\right)^2}{x^2+1}dx\)
6. I=\(\int\limits^5_1\frac{dx}{1+\sqrt{2x-1}}\)
7. I=\(\int\limits^3_1\frac{1+ln\left(x+1\right)}{x^2}dx\)
8.I=\(\int\limits^1_0\frac{x^3}{x^4+3x^2+2}dx\)
9. I=\(\int\limits^{\frac{\Pi}{4}}_0x\left(1+sin2x\right)dx\)
10. I=\(\int\limits^3_0\frac{x}{\sqrt{x+1}}dx\)
https://i.imgur.com/Pe6vPSJ.jpg
Tính tích phân: \(\int\limits^{log\left(1+\sqrt{2}\right)}_0\left(\dfrac{e^x-e^{-x}}{2}\right)^3\cdot\left(\dfrac{e^x+e^{-x}}{2}\right)^{11}dx\)
Tính các tích phân sau :
a) intlimits^1_0left(y^3+3y^2-2right)dy
b) intlimits^4_1left(t+dfrac{1}{sqrt{t}}-dfrac{1}{t^2}right)dt
c) intlimits^{dfrac{pi}{2}}_0left(2cos x-sin2xright)dx
d) intlimits^1_0left(3^s-2^sright)^2ds
e) intlimits^{dfrac{pi}{3}}_0cos3xdx+intlimits^{dfrac{3pi}{2}}_0cos3xdx+intlimits^{dfrac{5pi}{2}}_{dfrac{3pi}{2}}cos3xdx
g) intlimits^3_0left|x^2-x-2right|dx
h) intlimits^{dfrac{5pi}{4}}_{pi}dfrac{sin x-cos x}{sqrt{1+sin2x}}dx
i) intlimits^4_0dfrac{4x-1}{sqrt{2x+1}+2}dx
Đọc tiếp
Tính các tích phân sau :
a) \(\int\limits^1_0\left(y^3+3y^2-2\right)dy\)
b) \(\int\limits^4_1\left(t+\dfrac{1}{\sqrt{t}}-\dfrac{1}{t^2}\right)dt\)
c) \(\int\limits^{\dfrac{\pi}{2}}_0\left(2\cos x-\sin2x\right)dx\)
d) \(\int\limits^1_0\left(3^s-2^s\right)^2ds\)
e) \(\int\limits^{\dfrac{\pi}{3}}_0\cos3xdx+\int\limits^{\dfrac{3\pi}{2}}_0\cos3xdx+\int\limits^{\dfrac{5\pi}{2}}_{\dfrac{3\pi}{2}}\cos3xdx\)
g) \(\int\limits^3_0\left|x^2-x-2\right|dx\)
h) \(\int\limits^{\dfrac{5\pi}{4}}_{\pi}\dfrac{\sin x-\cos x}{\sqrt{1+\sin2x}}dx\)
i) \(\int\limits^4_0\dfrac{4x-1}{\sqrt{2x+1}+2}dx\)
Câu nào mình biết thì mình làm nha.
1) Đổi thành \(\dfrac{y^4}{4}+y^3-2y\) rồi thế số.KQ là \(\dfrac{-3}{4}\)
2) Biến đổi thành \(\dfrac{t^2}{2}+2\sqrt{t}+\dfrac{1}{t}\) và thế số.KQ là \(\dfrac{35}{4}\)
3) Biến đổi thành 2sinx + cos(2x)/2 và thế số.KQ là 1
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Tính các tích phân sau :
a) intlimits^1_0left(y-1right)^2sqrt{y}dy, đặt tsqrt{y}
b) intlimits^2_1left(x^2+1right)sqrt[3]{left(z-1right)^2}dz, đặt usqrt[3]{z-1}
c) intlimits^e_1dfrac{sqrt{4+5ln x}}{x}dx
d) intlimits^{dfrac{pi}{2}}_0left(cos^5varphi-sin^5varphiright)dvarphi
e) intlimits^{pi}_0cos^3alphacos3alpha dalpha
Đọc tiếp
Tính các tích phân sau :
a) \(\int\limits^1_0\left(y-1\right)^2\sqrt{y}dy\), đặt \(t=\sqrt{y}\)
b) \(\int\limits^2_1\left(x^2+1\right)\sqrt[3]{\left(z-1\right)^2}dz\), đặt \(u=\sqrt[3]{z-1}\)
c) \(\int\limits^e_1\dfrac{\sqrt{4+5\ln x}}{x}dx\)
d) \(\int\limits^{\dfrac{\pi}{2}}_0\left(\cos^5\varphi-\sin^5\varphi\right)d\varphi\)
e) \(\int\limits^{\pi}_0\cos^3\alpha\cos3\alpha d\alpha\)
Câu a)
Đặt \(y=\sqrt{t}\Rightarrow I_1=\int ^{1}_{0}(y-1)^2\sqrt{y}dy=\int ^{1}_{0}(t^2-1)^2td(t^2)\)
\(\Leftrightarrow I_1=2\int^{1}_{0}(t^2-1)^2t^2dt=2\int ^{1}_{0}(t^6-2t^4+t^2)dt\)
\(=2\left.\begin{matrix} 1\\ 0\end{matrix}\right|\left ( \frac{t^7}{7}-\frac{2t^5}{5}+\frac{t^3}{3} \right )=\frac{16}{105}\)
b) Đặt \(u=\sqrt[3]{z-1}\Rightarrow z=u^3+1\Rightarrow I_2=\int ^{1}_{0}[(u^3+1)^2+1]u^2d(u^3+1)\)
\(\Leftrightarrow I_2=3\int ^{1}_{0}[(u^3+1)^2+1]u^4du=3\int ^{1}_{0}(u^{10}+2u^7+2u^4)du\)
\(=3\left.\begin{matrix} 1\\ 0\end{matrix}\right|\left ( \frac{x^{11}}{11}+\frac{x^8}{4}+\frac{2x^5}{5} \right )=\frac{489}{220}\)
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c) Ta có:
\(I_3=\int ^{e}_{1}\frac{\sqrt{4+5\ln x}}{x}dx=\int ^{e}_{1}\sqrt{4+5\ln x}d(\ln x)\)
Đặt \(\sqrt{4+5\ln x}=t\Rightarrow I_3=\int ^{3}_{2}td\left (\frac{t^2-4}{5}\right)=\frac{2}{5}\int ^{3}_{2}t^2dt=\frac{38}{15}\)
d)
Xét \(\int ^{\frac{\pi}{2}}_{0}\cos ^5xdx=\int ^{\frac{\pi}{2}}_{0}\cos ^4xd(\sin x)=\int ^{\frac{\pi}{2}}_{0}(1-\sin ^2x)^2d(\sin x)\)
\(=\int ^{1}_{0}(1-t^2)^2dt\)
Xét \(\int ^{\frac{\pi}{2}}_{0}\sin ^5xdx=-\int ^{\frac{\pi}{2}}_{0}\sin ^4xd(\cos x)=-\int ^{\frac{\pi}{2}}_{0}(1-\cos ^2x)^2d(\cos x)=\int ^{1}_{0}(1-t^2)^2dt\)
Do đó \(\int ^{\frac{\pi}{2}}_{0}(\cos ^5x-\sin ^5x)dx=0\)
e)
Có \(\int \cos ^3x\cos 3xdx=\int \cos 3x\left ( \frac{3\cos x+\cos 3x}{4} \right )dx=\frac{1}{4}\int \cos ^23xdx+\frac{3}{4}\int \cos x\cos 3xdx\)
\(=\frac{1}{8}\int (1+\cos 6x)dx+\frac{3}{8}\int (\cos 4x+\cos 2x)dx\)
\(=\frac{1}{8}\int (1+\cos 6x)dx+\frac{3}{8}\int (\cos 4x+\cos 2x)dx=\frac{x}{8}+\frac{\sin 6x}{48}+\frac{3\sin 4x}{32}+\frac{3\sin 2x}{16}\)
Suy ra \(\int ^{\pi}_{0}\cos ^3x\cos 3xdx=\frac{\pi}{8}\)
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Tính các tích phân sau :
a) \(\int\limits^2_0\left|1-x\right|dx\)
b) \(\int\limits^{\dfrac{\pi}{2}}_0\sin^2xdx\)
c) \(\int\limits^{ln2}_0\dfrac{e^{2x+1}+1}{e^x}dx\)
d) \(\int\limits^{\pi}_0\sin2x\cos^2xdx\)
